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dkxd \(\hept{\begin{cases}\\\end{cases}}x-2=0;x+2=0\Leftrightarrow\hept{\begin{cases}\\\end{cases}x=+2;x=-2}\)
b/ \(\frac{x^2}{x^2-4}-\frac{x}{x+2}-\frac{2}{x-2}=\frac{x^2}{\left(x-2\right).\left(x+2\right)}-\frac{x.\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}-\frac{2.\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}\)
\(\frac{x^2-x^2-2x-2x+4}{\left(x-2\right).\left(x+2\right)}=\frac{4}{\left(x-2\right)\left(x+2\right)}\)
tới khúc này bí rồi ^^
a,ĐKXĐ của A là:\(x\ne+2;-2\)
b,\(\frac{x^2-x^2+2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)=\(\frac{4}{\left(x+2\right)\left(x-2\right)}\)
c,Để A\(\in\)Z=> (x+2)(x-2)\(\inƯ\)(4) hay \(x^2-4\inƯ\)(4)=\(\left(4;-4;2;-2;1;-1\right)\)
Ta có bảng
\(x^2-4\) | x |
4 | \(\sqrt{8}\) |
-4 | 0 |
2 | \(\sqrt{6}\) |
-2 | \(\sqrt{2}\) |
1 | \(\sqrt{5}\) |
Vậy A\(Z=>x\in\)( 0;\(\sqrt{8};\sqrt{6};\sqrt{2};\sqrt{5}\))
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(a\left(b^2+c^2+bc\right)+b\left(c^2+a^2+ac\right)+c\left(a^2+b^2+ab\right)\)
\(=ab^2+ac^2+abc+bc^2+a^2b+a^2c+b^2c+abc\)
\(=\left(ab^2+a^2b+abc\right)+\left(bc^2+b^2c+abc\right)+\left(ac^2+a^2c+abc\right)\)
\(=ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+b+c\right)\)
\(=\left(ab+bc+ac\right)\left(a+b+c\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(2x-2\right)^2-4x\left(x-2\right)+46\)
\(\Leftrightarrow A=4x^2-8x+4-4x^2+8x+46\)
\(\Leftrightarrow A=50\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(a+2\right)^2-\left(a+2\right)\left(a-2\right)\)
\(=\left(a+2\right)\left(a+2-a+2\right)\)
\(=\left(a+2\right)\cdot4=4a+8\)